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linear elastic fluid model with attenuation
Posted Jan 21, 2016, 5:16 a.m. EST Acoustics & Vibrations Version 5.0 1 Reply
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Hi everyone,
I'm developing a model to predict the sound pressure around a point source. All the domains are composed by water and I need a fluid model with attenuation. So i set up the attenuation coefficient expressed as Np/m as function of frequency (I'm dealing with frequency domain). The trouble is that the sound pressure calculated through the model with attenuation results bigger then the sound pressure calculated with a linear elastic model without attenuation. the only way a found to solve the problem was to put a minus sign forward the attenuation coefficient, but i think this solution has no physic explaination. Can someone help me to understand the problem?
I'm developing a model to predict the sound pressure around a point source. All the domains are composed by water and I need a fluid model with attenuation. So i set up the attenuation coefficient expressed as Np/m as function of frequency (I'm dealing with frequency domain). The trouble is that the sound pressure calculated through the model with attenuation results bigger then the sound pressure calculated with a linear elastic model without attenuation. the only way a found to solve the problem was to put a minus sign forward the attenuation coefficient, but i think this solution has no physic explaination. Can someone help me to understand the problem?
1 Reply Last Post Jan 25, 2016, 1:01 p.m. EST
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Posted:
9 years ago
Hi Pietro,
One possible reason that comes to mind is that the attenuation (and other forms of damping) also affect the natural frequencies. So, for example, if there is a natural frequency at 100 Hz without attenuation and that drops to say 90 Hz with attenuation, and you request a solution at a frequency of 90 Hz the resulting pressure may be higher with attenuation.
Nagi Elabbasi
Veryst Engineering
One possible reason that comes to mind is that the attenuation (and other forms of damping) also affect the natural frequencies. So, for example, if there is a natural frequency at 100 Hz without attenuation and that drops to say 90 Hz with attenuation, and you request a solution at a frequency of 90 Hz the resulting pressure may be higher with attenuation.
Nagi Elabbasi
Veryst Engineering